How many milliliters of NH3 (at STP) are needed to react with 45.0 mL of NO2 (at STP) according to the equation:

8NO2 (g) + 6NH3 )g) yield 7N2O (g) + 9 H2O.

How to set this up?????

Thanks.

The answer that I got was 33.8---is this correct.

Yes. I obtained 33.750 which rounds to 33.8 mL to three significant figures.

To set up this problem, you need to determine the stoichiometry of the reaction between NH3 and NO2. The coefficients in the balanced equation (8NO2 + 6NH3 → 7N2O + 9H2O) give you the molar ratio between the reactants and products.

1. Start by converting the given volume of NO2 to moles. You can use the ideal gas law at standard temperature and pressure (STP: 0 degrees Celsius and 1 atmosphere pressure) to convert volume to moles. The molar volume of any gas at STP is 22.4 L/mol.

45.0 mL NO2 × (1 L / 1000 mL) × (1 mol / 22.4 L) = X mol NO2

2. Use the mole ratio between NH3 and NO2 to determine the number of moles of NH3 needed. From the balanced equation, you can see that 6 moles of NH3 react with 8 moles of NO2.

X mol NO2 × (6 mol NH3 / 8 mol NO2) = Y mol NH3

3. Finally, convert the number of moles of NH3 to milliliters using the molar volume of gas at STP.

Y mol NH3 × (22.4 L/mol) × (1000 mL / 1 L) = Z mL NH3

The final value, Z, will give you the number of milliliters of NH3 needed to react with 45.0 mL of NO2 at STP according to the given balanced equation.